Question

A vector makes an angles with x y and z axes that are in the ratio 1:2:3 respectively. The angle made by the vector with y axis is

Answer 1

Let angles with x , y and z axes are a , b and c respectively.

a/c to question,

a : b : c = 1 : 2 : 3

so we can assume , a = k, b = 2k and c = 3k

but we know, direction of cosine ,

cos²a + cos²b + cos²c = 1

or, cos²k + cos²(2k) + cos²(3k) = 1

we know, (1 + cos2α)/2 = cos²α

so, (1 + cos2k)/2 + (1 + cos4k)/2 + (1 + cos6k)/2 = 1

or, cos2k + cos4k + cos6k = -1

Let p = 2k then, cosp + cos2p + cos3p = -1

using formula, cos2α = 2cos²α - 1

cos3α = 4cos³α - 3cosα

so, cosp + (2cos² p - 1) + (4cos³p - 3cosp) = -1

or, cosp + 2cos²p + 4cos³p - 3cosp = 0

or, 4cos³p + 2cos²p - 2cos²p = 0

or, 2cosp(2cos²p + cosp - 1) = 0

or, cosp (cosp + 1)(2cosp - 1) = 0

or, cosp = 0, -1 or 1/2

so, cos2k = cos(π/2), cos(π) or, cos(π/3)

2k = π/2, π, or π/3

k = π/4, π/2 or, π/6

case 1 : when , k = π/4 , b = 2k = π/2

case 2 : when , k = π/2, b = 2k = π

case 3 : when , k = π/6 , b = 2k = π/3

Answer 2

Answer45*

Explanation:

Let angles with x , y and z axes are a , b and c respectively.

a/c to question,

a : b : c = 1 : 2 : 3

so we can assume , a = k, b = 2k and c = 3k

but we know, direction of cosine ,

cos²a + cos²b + cos²c = 1

or, cos²k + cos²(2k) + cos²(3k) = 1

we know, (1 + cos2α)/2 = cos²α

so, (1 + cos2k)/2 + (1 + cos4k)/2 + (1 + cos6k)/2 = 1

or, cos2k + cos4k + cos6k = -1

Let p = 2k then, cosp + cos2p + cos3p = -1

using formula, cos2α = 2cos²α - 1

cos3α = 4cos³α - 3cosα

so, cosp + (2cos² p - 1) + (4cos³p - 3cosp) = -1

or, cosp + 2cos²p + 4cos³p - 3cosp = 0

or, 4cos³p + 2cos²p - 2cos²p = 0

or, 2cosp(2cos²p + cosp - 1) = 0

or, cosp (cosp + 1)(2cosp - 1) = 0

or, cosp = 0, -1 or 1/2

so, cos2k = cos(π/2), cos(π) or, cos(π/3)

2k = π/2, π, or π/3

k = π/4, π/2 or, π/6

case 1 : when , k = π/4 , b = 2k = π/2

case 2 : when , k = π/2, b = 2k = π

case 3 : when , k = π/6 , b = 2k = π/3